We are learning compareTo() method and equals() method, but I am unsure how to properly print the words in alphabetical order.
I have 2 Strings:
String firstString;
String secondString;
that are assigned values (elsewhere).
I have tried:
System.out.println(firstString.CompareToIgnoreCase(secondString));
but am only getting integers in my output as opposed to full words.
As this is homework, I'll only address your stumbling block.
As per the documentation, compareToIgnoreCase() (and more generally, compareTo()) returns an integer, either negative, zero or positive, depending on whether the first parameter's order is less than, the same or greater respectively than the second parameter.
You'll need to check the result of compareToIgnoreCase(), then print based on that, for example:
if (firstString.compareToIgnoreCase(secondString) > 0) {
// do something
} else {
// do something else
}
For those taking the same class, this solves the requirements for the question:
if (firstString.compareTo(secondString) < 0) {
System.out.println(firstString + " " + secondString);
}
else {
System.out.println(secondString + " " + firstString);
}
I was wondering how I should write this if statement in my assignment.
The question is:
Print option 2 if one of the strings begins with the letter w or has 5 characters.
I would use the .contains to find the "w".
if (two.contains("w") {
System.out.println(two);
but the one with the characters I am unsure how to find the method.
If you have a List or Set, you may need to loop over them one by one, to do the actual comparing try this:
(two.startsWith("w") || two.length() == 5) {
System.out.println(two);
}
The first condition checks if given String object starts with given char, and the other one counts the number of characters in the String and checks your desired lenght, which is 5.
For more useful information and String object has, check this out String (Java Platform SE 7)
String aString = ".....";
if (aString.startsWith("w") || aString.length() == 5) {
System.out.println("option 2");
}
The method .contains returns true if find a substring (in this case "w") in string. You should use .startsWith method.
Thanks for checking out my question.
Starting off, the program has the following goal; the user inputs currency formatted as "xD xC xP xH"; the program checks the input is correct and then prints back the 'long' version: "x Dollars, x Cents, x Penny's, x half penny's"
Here I have some code that takes input from user as String currencyIn, splits the string into array tokens, then replaces the D's with Dollars etc and prints the output.
public class parseArray
{
public parseArray()
{
System.out.print('\u000c');
String CurrencyFormat = "xD xS xP xH";
System.out.println("Please enter currency in the following format: \""+CurrencyFormat+"\" where x is any integer");
System.out.println("\nPlease take care to use the correct spacing enter the exact integer plus type of coin\n\n");
Scanner input = new Scanner(System.in);
String currencyIn = input.nextLine();
currencyIn.toUpperCase();
System.out.println("This is the currency you entered: "+currencyIn);
String[] tokens = currencyIn.split(" ");
for (String t : tokens)
{
System.out.println(t);
}
String dollars = tokens[0].replaceAll("D", " Dollars ");
String cents = tokens[1].replaceAll("C", " cents");
String penny = tokens[2].replaceAll("P", " Penny's");
String hPenny = tokens[3].replaceAll("H", " Half penny's");
System.out.println(" "+dollars+ " " +cents+ " " +penny+ " " +hPenny);
input.close();
}
}
Question 1: At the moment the program prints out pretty anything you put in. how do I establish some input control? I've seen this done in textbooks with switch statement and a series of if statements, but were too complicated for me. Would it parse characters using charAt() for each element of the array?
Question 2: Is there a 'better' way to print the output? My friend said converting my 4 strings (dollars, cents, penny's, hpenny's) into elements 0, 1, 2, 3 of a new array (called newArray) and print like this:
System.out.println(Arrays.toString(newArray));
Many thanks in advance.
There is a neat solution, involving Regular Expressions, Streams and some lambdas. Core concept is that we define the input format through a regular expression. We need some sequence of digits, followed by a 'D' or a 'd', followed by a " ", followed by a sequence of digits, followed by a C or c,... I will skip derivation of this pattern, it is explained in the regular expression tutorial I linked above. We will find that
final String regex = "([0-9]+)[D|d]\\ ([0-9]+)[C|c]\\ ([0-9]+)[P|p]\\ ([0-9]+)[H|h]";
satisfies our needs. With this regular expression we can now determine whether our input String has the right format (input.matches(regex)), as well as extract the bits of information we are actually interested in (input.replaceAll(regex, "$1 $2 $3 $4"). Sadly, replaceAll yields another String, but it will contain the four digit sequences we are interested in, divided by a " ". We will use some stream-magic to transform this String into a long[] (where the first cell holds the D-value, the second holds the C-value,...). The final program looks like this:
import java.util.Arrays;
public class Test {
public static void main(String... args) {
final String input = args[0];
final String regex =
"([0-9]+)[D|d]\\ ([0-9]+)[C|c]\\ ([0-9]+)[P|p]\\ ([0-9]+)[H|h]";
if (input.matches(regex) == false) {
throw new IllegalArgumentException("Input is malformed.");
}
long[] values = Arrays.stream(input.replaceAll(regex, "$1 $2 $3 $4").split(" "))
.mapToLong(Long::parseLong)
.toArray();
System.out.println(Arrays.toString(values));
}
}
If you want to have a List<Long> instead a long[] (or a List<Integer> instead of an int[]), you would use
List<Long> values = Arrays.stream(input.replaceAll(regex, "$1 $2 $3 $4").split(" "))
.map(Long::parseLong)
.collect(Collectors.toList());
It is necessary to change mapToLong to map to receive a Stream<Long> instead of a LongStream. I am sure that one could somehow write a custom Collector for LongStream to transform it into a List<Long>, but I found this solution more readable and reliable (after all, the Collector used comes from Oracle, I trust they test their code extensively).
Here is some example call:
$> java Test "10D 9c 8p 7H"
[10, 9, 8, 7]
$> java Test "10E 9C 8P 7H"
Exception in thread "main" java.lang.IllegalArgumentException: Input is malformed.
at Test.main(Test.java:10)
$> java Test "10D 9C 8P 7H 10D 9C 8P 7H"
Exception in thread "main" java.lang.IllegalArgumentException: Input is malformed.
at Test.main(Test.java:10)
Question1
You can actually check if the input is what it's supposed to be with simple checks. For example, you can check the first element like this:
if(tokens[0].charAt(1).equals("D"))
return true;
else
return false;
Another way to check if the input is correct is by using Regular Expressions, but I assume you are a beginner and this is too much trouble for you, although it is the better way. So I leave it to you to look through it later.
Question2
You can actually listen to your friend and do as they said. You can write it as follows:
for(int i = 0; i < 4; i++)
System.out.print(" " + tokens[i])
System.out.println();
Or you may use
System.out.println(Arrays.toString(newArray));
And you have saved newArray like this:
newArray[0] = " " + tokens[0];
you could use the .equals() method to see if what a user has typed in matches what you have
if (currencyIn.equals("CurrencyFormat"))
{
...
}
this is probably the simplest way i can think of!
I am trying to concatenate strings in Java. Why isn't this working?
public class StackOverflowTest {
public static void main(String args[]) {
int theNumber = 42;
System.out.println("Your number is " . theNumber . "!");
}
}
You can concatenate Strings using the + operator:
System.out.println("Your number is " + theNumber + "!");
theNumber is implicitly converted to the String "42".
The concatenation operator in java is +, not .
Read this (including all subsections) before you start. Try to stop thinking the php way ;)
To broaden your view on using strings in Java - the + operator for strings is actually transformed (by the compiler) into something similar to:
new StringBuilder().append("firstString").append("secondString").toString()
There are two basic answers to this question:
[simple] Use the + operator (string concatenation). "your number is" + theNumber + "!" (as noted elsewhere)
[less simple]: Use StringBuilder (or StringBuffer).
StringBuilder value;
value.append("your number is");
value.append(theNumber);
value.append("!");
value.toString();
I recommend against stacking operations like this:
new StringBuilder().append("I").append("like to write").append("confusing code");
Edit: starting in java 5 the string concatenation operator is translated into StringBuilder calls by the compiler. Because of this, both methods above are equal.
Note: Spaceisavaluablecommodity,asthissentancedemonstrates.
Caveat: Example 1 below generates multiple StringBuilder instances and is less efficient than example 2 below
Example 1
String Blam = one + two;
Blam += three + four;
Blam += five + six;
Example 2
String Blam = one + two + three + four + five + six;
Out of the box you have 3 ways to inject the value of a variable into a String as you try to achieve:
1. The simplest way
You can simply use the operator + between a String and any object or primitive type, it will automatically concatenate the String and
In case of an object, the value of String.valueOf(obj) corresponding to the String "null" if obj is null otherwise the value of obj.toString().
In case of a primitive type, the equivalent of String.valueOf(<primitive-type>).
Example with a non null object:
Integer theNumber = 42;
System.out.println("Your number is " + theNumber + "!");
Output:
Your number is 42!
Example with a null object:
Integer theNumber = null;
System.out.println("Your number is " + theNumber + "!");
Output:
Your number is null!
Example with a primitive type:
int theNumber = 42;
System.out.println("Your number is " + theNumber + "!");
Output:
Your number is 42!
2. The explicit way and potentially the most efficient one
You can use StringBuilder (or StringBuffer the thread-safe outdated counterpart) to build your String using the append methods.
Example:
int theNumber = 42;
StringBuilder buffer = new StringBuilder()
.append("Your number is ").append(theNumber).append('!');
System.out.println(buffer.toString()); // or simply System.out.println(buffer)
Output:
Your number is 42!
Behind the scene, this is actually how recent java compilers convert all the String concatenations done with the operator +, the only difference with the previous way is that you have the full control.
Indeed, the compilers will use the default constructor so the default capacity (16) as they have no idea what would be the final length of the String to build, which means that if the final length is greater than 16, the capacity will be necessarily extended which has price in term of performances.
So if you know in advance that the size of your final String will be greater than 16, it will be much more efficient to use this approach to provide a better initial capacity. For instance, in our example we create a String whose length is greater than 16, so for better performances it should be rewritten as next:
Example optimized :
int theNumber = 42;
StringBuilder buffer = new StringBuilder(18)
.append("Your number is ").append(theNumber).append('!');
System.out.println(buffer)
Output:
Your number is 42!
3. The most readable way
You can use the methods String.format(locale, format, args) or String.format(format, args) that both rely on a Formatter to build your String. This allows you to specify the format of your final String by using place holders that will be replaced by the value of the arguments.
Example:
int theNumber = 42;
System.out.println(String.format("Your number is %d!", theNumber));
// Or if we need to print only we can use printf
System.out.printf("Your number is still %d with printf!%n", theNumber);
Output:
Your number is 42!
Your number is still 42 with printf!
The most interesting aspect with this approach is the fact that we have a clear idea of what will be the final String because it is much more easy to read so it is much more easy to maintain.
The java 8 way:
StringJoiner sj1 = new StringJoiner(", ");
String joined = sj1.add("one").add("two").toString();
// one, two
System.out.println(joined);
StringJoiner sj2 = new StringJoiner(", ","{", "}");
String joined2 = sj2.add("Jake").add("John").add("Carl").toString();
// {Jake, John, Carl}
System.out.println(joined2);
You must be a PHP programmer.
Use a + sign.
System.out.println("Your number is " + theNumber + "!");
"+" instead of "."
Use + for string concatenation.
"Your number is " + theNumber + "!"
This should work
public class StackOverflowTest
{
public static void main(String args[])
{
int theNumber = 42;
System.out.println("Your number is " + theNumber + "!");
}
}
For exact concatenation operation of two string please use:
file_names = file_names.concat(file_names1);
In your case use + instead of .
For better performance use str1.concat(str2) where str1 and str2 are string variables.
String.join( delimiter , stringA , stringB , … )
As of Java 8 and later, we can use String.join.
Caveat: You must pass all String or CharSequence objects. So your int variable 42 does not work directly. One alternative is using an object rather than primitive, and then calling toString.
Integer theNumber = 42;
String output =
String // `String` class in Java 8 and later gained the new `join` method.
.join( // Static method on the `String` class.
"" , // Delimiter.
"Your number is " , theNumber.toString() , "!" ) ; // A series of `String` or `CharSequence` objects that you want to join.
) // Returns a `String` object of all the objects joined together separated by the delimiter.
;
Dump to console.
System.out.println( output ) ;
See this code run live at IdeOne.com.
In java concatenate symbol is "+".
If you are trying to concatenate two or three strings while using jdbc then use this:
String u = t1.getString();
String v = t2.getString();
String w = t3.getString();
String X = u + "" + v + "" + w;
st.setString(1, X);
Here "" is used for space only.
In Java, the concatenation symbol is "+", not ".".
"+" not "."
But be careful with String concatenation. Here's a link introducing some thoughts from IBM DeveloperWorks.
You can concatenate Strings using the + operator:
String a="hello ";
String b="world.";
System.out.println(a+b);
Output:
hello world.
That's it
So from the able answer's you might have got the answer for why your snippet is not working. Now I'll add my suggestions on how to do it effectively. This article is a good place where the author speaks about different way to concatenate the string and also given the time comparison results between various results.
Different ways by which Strings could be concatenated in Java
By using + operator (20 + "")
By using concat method in String class
Using StringBuffer
By using StringBuilder
Method 1:
This is a non-recommended way of doing. Why? When you use it with integers and characters you should be explicitly very conscious of transforming the integer to toString() before appending the string or else it would treat the characters to ASCI int's and would perform addition on the top.
String temp = "" + 200 + 'B';
//This is translated internally into,
new StringBuilder().append( "" ).append( 200 ).append('B').toString();
Method 2:
This is the inner concat method's implementation
public String concat(String str) {
int olen = str.length();
if (olen == 0) {
return this;
}
if (coder() == str.coder()) {
byte[] val = this.value;
byte[] oval = str.value;
int len = val.length + oval.length;
byte[] buf = Arrays.copyOf(val, len);
System.arraycopy(oval, 0, buf, val.length, oval.length);
return new String(buf, coder);
}
int len = length();
byte[] buf = StringUTF16.newBytesFor(len + olen);
getBytes(buf, 0, UTF16);
str.getBytes(buf, len, UTF16);
return new String(buf, UTF16);
}
This creates a new buffer each time and copies the old content to the newly allocated buffer. So, this is would be too slow when you do it on more Strings.
Method 3:
This is thread safe and comparatively fast compared to (1) and (2). This uses StringBuilder internally and when it allocates new memory for the buffer (say it's current size is 10) it would increment it's 2*size + 2 (which is 22). So when the array becomes bigger and bigger this would really perform better as it need not allocate buffer size each and every time for every append call.
private int newCapacity(int minCapacity) {
// overflow-conscious code
int oldCapacity = value.length >> coder;
int newCapacity = (oldCapacity << 1) + 2;
if (newCapacity - minCapacity < 0) {
newCapacity = minCapacity;
}
int SAFE_BOUND = MAX_ARRAY_SIZE >> coder;
return (newCapacity <= 0 || SAFE_BOUND - newCapacity < 0)
? hugeCapacity(minCapacity)
: newCapacity;
}
private int hugeCapacity(int minCapacity) {
int SAFE_BOUND = MAX_ARRAY_SIZE >> coder;
int UNSAFE_BOUND = Integer.MAX_VALUE >> coder;
if (UNSAFE_BOUND - minCapacity < 0) { // overflow
throw new OutOfMemoryError();
}
return (minCapacity > SAFE_BOUND)
? minCapacity : SAFE_BOUND;
}
Method 4
StringBuilder would be the fastest one for String concatenation since it's not thread safe. Unless you are very sure that your class which uses this is single ton I would highly recommend not to use this one.
In short, use StringBuffer until you are not sure that your code could be used by multiple threads. If you are damn sure, that your class is singleton then go ahead with StringBuilder for concatenation.
First method: You could use "+" sign for concatenating strings, but this always happens in print.
Another way: The String class includes a method for concatenating two strings: string1.concat(string2);
import com.google.common.base.Joiner;
String delimiter = "";
Joiner.on(delimiter).join(Lists.newArrayList("Your number is ", 47, "!"));
This may be overkill to answer the op's question, but it is good to know about for more complex join operations. This stackoverflow question ranks highly in general google searches in this area, so good to know.
you can use stringbuffer, stringbuilder, and as everyone before me mentioned, "+". I'm not sure how fast "+" is (I think it is the fastest for shorter strings), but for longer I think builder and buffer are about equal (builder is slightly faster because it's not synchronized).
here is an example to read and concatenate 2 string without using 3rd variable:
public class Demo {
public static void main(String args[]) throws Exception {
InputStreamReader r=new InputStreamReader(System.in);
BufferedReader br = new BufferedReader(r);
System.out.println("enter your first string");
String str1 = br.readLine();
System.out.println("enter your second string");
String str2 = br.readLine();
System.out.println("concatenated string is:" + str1 + str2);
}
}
There are multiple ways to do so, but Oracle and IBM say that using +, is a bad practice, because essentially every time you concatenate String, you end up creating additional objects in memory. It will utilize extra space in JVM, and your program may be out of space, or slow down.
Using StringBuilder or StringBuffer is best way to go with it. Please look at Nicolas Fillato's comment above for example related to StringBuffer.
String first = "I eat"; String second = "all the rats.";
System.out.println(first+second);
Using "+" symbol u can concatenate strings.
String a="I";
String b="Love.";
String c="Java.";
System.out.println(a+b+c);
The user enters an array of numbers such as "123456789"
and then the user is supposed to enter a string of commands such as "PE"
and my program will check each character in the string.
P = Remove the last number entered, which is 9
E = Prints out the final result.
To this point everything was working fine.
But my problem is that I have to be able to read the command when the user inputs "P(any number)", and I have to place that number at the end of the array of numbers.
So if it was P(4), the result would be "1234567894".
How can I do this ?
This is what I have right now for P and E.
if (commandList.get(x).equals("P(")) {
JOptionPane.showMessageDialog(null, "HI");
}
else if (commandList.get(x).equals("P")) {
strList.remove(strList.size()-1);
x++;
}
As far as I understand, your problem is to extract the 'digit' from the string "P(digit)" ?
Well you have to parse the input string, e.g., by searching for bounding characters, here '(' and ')'. For example by something like that:
String s = "P(3)";
String number = s.substring(s.indexOf("(") + 1, s.indexOf(")"));
strList.append(number);
But this would also accept anything within the bracktes, e.g., "foooo" from "P(foooo)", and it would crash if ')' cannot be found, e.g., s = "P(123", and it would accept "P(3)blah54" which is strictly spoken invalid. To improve your parser, you can use regular expressions, e.g., to check if the input is exactly of the form "P(2)" for exactly a single digit, you can use
String s = "P(3)";
if (!s.matches("P\\(\\d\\)"))
throw new RuntimeException("invalid input format: " + s);
String number = s.substring(s.indexOf("(") + 1, s.indexOf(")"));
strList.append(number);
If you explicitly want to cast the digit to an int, you might have a look at Integer.parseInt(...), e.g.,
int i = Integer.parseInt(number);